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Question

Bullet wait is 0.05 kg penetrates into a block of food of mass 2 kg velocity of 150 M per second hand gets embedded in the block of wood if the block of wood was initially at rest calculate the work done by the system

Answer 1

If by "work done by the system" you mean the work done on the bullet, then the work is just the difference between final and initial kinetic energies of the bullet.  If you mean something else, please clarify.  I will work the problem assuming you mean the former.  (Also, please proofread your question.  It says "wait" instead of "weight", "food" instead of "wood", "hand" instead of "and", ...)

 

m = mass of bullet = 0.05 kg

M = mass of wood block = 2 kg

v₀ = initial speed of bullet = 150 m/s

 

v = final speed of bullet = ?

 

Conservation of linear momentum:

 

mv₀ + M(0) = (m + M)v

 

v = mv₀/(m + M) = (0.05)(150)/(0.05 + 2) m/s = 3.659 m/s

 

K₀ = (1/2)mv₀2

K = (1/2)mv²

 

W = ΔK = K - K₀ = (m/2)(v² - v₀²) = (0.05/2)(3.659² - 150²) J = -562.2 J

 

The negative sign means the system did work on the bullet.